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how to fit exponential distribution function on data?

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Mos_bad on 25 Jan 2020 at 19:31
Answered: Image Analyst on 26 Jan 2020 at 3:57
The vector m follows the truncated exponential equation (F_M) and it is shown by solid black line in figure. I intend to fit an exponential distribution function to data and find the parameter lambda (1/mean). Even though I've used fitdist(x,distname), the fitted exp. dist. shown in dashed line which is way different from the data. here is the code:
M_min=4.5; M_max=8.0;
a=4.56; b=1.0;
F_M=(1-exp(-beta*(m-M_min))) / (1-(exp(-beta*(M_max-M_min)))); % CDF of Mag.
pd = fitdist(m','Exponential');
figure(1); plot(m,1-F_M,'-','linewidth',2);
hold on; plot(m,1-cdf(pd,m),'--');
legend('data','fitted dist')


Walter Roberson
Walter Roberson on 25 Jan 2020 at 23:53
a and b are not defined in the third line.
Mos_bad on 25 Jan 2020 at 23:57
just editted. thanks for pointing that out.

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Answers (2)

Walter Roberson
Walter Roberson on 26 Jan 2020 at 0:02
You do not have an exponential distribution. (1 minus an exponential) is not an exponential.
On the other hand if you fit using the equation
instead of
then you get pretty much a perfect fit.

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Mos_bad on 26 Jan 2020 at 0:38
All I want to do is to devide vertical axis to 1000 intervals and pick a random value of magnitude (horizental axis) at each interval. Kind of Latin hypercube sampling.

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